Question

solve for x.
$3x^2 + 22x + 24 > 4x$
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Explanation:

Step1: Rearrange the inequality

Subtract \(4x\) from both sides to get the quadratic inequality in standard form:
\(3x^{2}+22x - 4x+24>0\)
Simplify to \(3x^{2}+18x + 24>0\)
We can factor out a 3: \(3(x^{2}+6x + 8)>0\)
Then factor the quadratic inside the parentheses: \(3(x + 2)(x+4)>0\)

Step2: Find critical points

Set each factor equal to zero:
\(x+2 = 0\) gives \(x=-2\)
\(x + 4=0\) gives \(x=-4\)

Step3: Test intervals

The critical points divide the number line into three intervals: \((-\infty,-4)\), \((-4,-2)\), and \((-2,\infty)\)

• For \(x < - 4\) (e.g., \(x=-5\)): \(3(-5 + 2)(-5+4)=3(-3)(-1)=9>0\), so the inequality holds here.
• For \(-4 < x<-2\) (e.g., \(x = - 3\)): \(3(-3 + 2)(-3 + 4)=3(-1)(1)=-3<0\), so the inequality does not hold here.
• For \(x>-2\) (e.g., \(x = 0\)): \(3(0 + 2)(0 + 4)=3(2)(4)=24>0\), so the inequality holds here.

Answer:

\(x < - 4\) or \(x>-2\)